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Chapter 3: Problem 4

If \(\sum x_{n}\) and \(\sum y_{n}\) are convergent, show that \(\sum\left(x_{n}+y_{n}\right)\) is convergent.

Short Answer

Expert verified
The sum of two convergent series \(\sum x_{n}\) and \(\sum y_{n}\) is also a convergent series.

Step by step solution

01

Convergence of \(\sum x_{n}\) and \(\sum y_{n}\)

The question mentions that \(\sum x_{n}\) and \(\sum y_{n}\) are convergent. By the definition of convergence, for these series, the partial sums \(S_{N}^{x}=\sum_{n=1}^{N} x_{n}\) and \(S_{N}^{y}=\sum_{n=1}^{N} y_{n}\) become more and more steady as \(N\) approaches infinity.
02

Sum of the series

We define the sequence \(z_{n}=x_{n}+y_{n}\), and we consider the infinite series \(\sum z_{n} = \sum (x_{n} + y_{n})\). The partial sum of this series at \(N\) terms is given by \(S_{N}^{z}=\sum_{n=1}^{N} z_{n} = \sum_{n=1}^{N} (x_{n} + y_{n}) = \sum_{n=1}^{N} x_{n} + \sum_{n=1}^{N} y_{n} = S_{N}^{x} + S_{N}^{y}\).
03

Show Convergence of the Sum

As \(N\) approaches infinity, both \(S_{N}^{x}\) and \(S_{N}^{y}\) approach a steady state (because \(\sum x_{n}\) and \(\sum y_{n}\) are convergent). Hence, their sum \(S_{N}^{z} = S_{N}^{x} + S_{N}^{y}\) also approaches a steady state as \(N\) approaches infinity. Therefore, \(\sum z_{n} = \sum (x_{n} + y_{n})\) is convergent.

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Most popular questions from this chapter

Establish the convergence and find the limits of the following sequences: (a) \(\left(\left(1+1 / n^{2}\right)^{n^{2}}\right)\). (b) \(\left((1+1 / 2 n)^{n}\right)\), (c) \(\left(\left(1+1 / n^{2}\right)^{2 n^{2}}\right)\). (d) \(\left((1+2 / n)^{n}\right)\).

Show that if \(\left(x_{n}\right)\) is unbounded, then there exists a subsequence \(\left(x_{n_{k}}\right)\) such that \(\lim \left(1 / x_{n_{2}}\right)=0\).

Let \(b \in \mathbb{R}\) satisfy \(0

If \(\sum a_{n}\) with \(a_{n}>0\) is convergent, then is \(\sum \sqrt{a_{n}}\) always convergent? Either prove it or give a counterexample.

Show that \(\lim \left(\frac{1}{n}-\frac{1}{n+1}\right)=0\).
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